Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter7: Equations And Inequalities In Two Variables
Section7.3: Distance And Slope
Problem 68PS
Related questions
Question
![**Problem Statement:**
The tenth term of an arithmetic sequence is \(\frac{73}{2}\) and the second term is \(\frac{9}{2}\). Find the first term.
**Solution:**
To find the first term of an arithmetic sequence \(a_1\), we need to use the given information about the terms of the sequence.
1. **Identify given terms:**
- 10th term (a₁₀) = \(\frac{73}{2}\)
- 2nd term (a₂) = \(\frac{9}{2}\)
2. **Use the arithmetic sequence formula:**
The \(n\)-th term of an arithmetic sequence can be written as:
\[
a_n = a_1 + (n-1)d
\]
where \(a_n\) is the \(n\)-th term, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number.
3. **Set up equations:**
Using the information provided:
For \(a_2\) (second term):
\[
a_2 = a_1 + d
\]
Substitute the given value for \(a_2\):
\[
\frac{9}{2} = a_1 + d \quad ...(1)
\]
For \(a_{10}\) (tenth term):
\[
a_{10} = a_1 + 9d
\]
Substitute the given value for \(a_{10}\):
\[
\frac{73}{2} = a_1 + 9d \quad ...(2)
\]
4. **Solve the system of equations:**
Subtract equation (1) from equation (2):
\[
\frac{73}{2} - \frac{9}{2} = (a_1 + 9d) - (a_1 + d)
\]
Simplify:
\[
\frac{64}{2} = 8d
\]
\[
32 = 8d
\]
\[
d = 4
\]
5. **Substitute \(d\) back into equation (1):](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f29774d-39d7-4c90-90ba-0f0924899c12%2Fa81ecd8f-57a9-4249-8d47-c6fa3a84067d%2F4ou6oh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The tenth term of an arithmetic sequence is \(\frac{73}{2}\) and the second term is \(\frac{9}{2}\). Find the first term.
**Solution:**
To find the first term of an arithmetic sequence \(a_1\), we need to use the given information about the terms of the sequence.
1. **Identify given terms:**
- 10th term (a₁₀) = \(\frac{73}{2}\)
- 2nd term (a₂) = \(\frac{9}{2}\)
2. **Use the arithmetic sequence formula:**
The \(n\)-th term of an arithmetic sequence can be written as:
\[
a_n = a_1 + (n-1)d
\]
where \(a_n\) is the \(n\)-th term, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number.
3. **Set up equations:**
Using the information provided:
For \(a_2\) (second term):
\[
a_2 = a_1 + d
\]
Substitute the given value for \(a_2\):
\[
\frac{9}{2} = a_1 + d \quad ...(1)
\]
For \(a_{10}\) (tenth term):
\[
a_{10} = a_1 + 9d
\]
Substitute the given value for \(a_{10}\):
\[
\frac{73}{2} = a_1 + 9d \quad ...(2)
\]
4. **Solve the system of equations:**
Subtract equation (1) from equation (2):
\[
\frac{73}{2} - \frac{9}{2} = (a_1 + 9d) - (a_1 + d)
\]
Simplify:
\[
\frac{64}{2} = 8d
\]
\[
32 = 8d
\]
\[
d = 4
\]
5. **Substitute \(d\) back into equation (1):
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